Question

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 420 grams and a standard deviation of 24 grams. Find the weight that corresponds to each event. (Use Excel or Appendix C to calculate the z-value. Round your final answers to 2 decimal places.) a. Highest 20 percent b. Middle 60 percent to c. Highest 80 percent d. Lowest 15 percent.

Answer 1

Answer and Step-by-step explanation:

Round your final answers to 2 decimal places

μ = 420

σ = 24

a. Highest 20 percent

P (Z > a-420/24) = 0.20

For a highest 20%, according to the table attached, we have to find the z score for 0.30 (0.5 - 0.2)

P(0 < Z < 0.84) = 0.3

a-420/24 = 0.84

a-420 = 20.16

a = 20.16 + 420 = 440.16

weight 440.16

b. Middle 60 percent

P (0 < Z < b-420/24) = 0.3

b = 440.16

P (c-420/24 < Z < 0) = 0.3

c-420/24 = -0.84

c-420 = -20.16

c = 399.84

Weight from 399.84 to 440.16

c. Highest 80 percent

P (d-420/24 < Z < 0) = 0.3

d-420/24 = -0.84

d-420 = -20.16

d = 399.84

d. Lowest 15 percent

P(e-420/24 < Z < 0) = 0.35

e-420/24 = -1.04

e-420 = -29.76

e = 390.24

Answer 2

Probability of an event is the chance of occurrence of that event.

The weight corresponding to each event is given as:

• a) Highest 20% : X > 440.4 grams,
• b Middle 60% : 399.4 < X < 440.4 grams
• c) Highest 80% :  X > 399.4 grams
• d) Lowest 15% :  X < 395.16 grams

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have $X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write $P(Z \leq z) = P(Z < z)$ )

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = p \: value$

For the given case, let we have

X = random variable tracking the weights of small Starbucks coffee

Then, by the given data, we have:

$X \sim N(420, 24)$

Evaluating the weights for the given cases:

• a) Highest 20%

Let the value be x such that $(X > x) = 20\% \: of \: values$

Then, converting it to z score, we get:

$P(Z > \dfrac{x - 420}{24}) = 0.2 = 1 - P( Z \leq \dfrac{x - 420}{24} )\\\\P( Z \leq \dfrac{x - 420}{24}) = 1 - 0.2 = 0.8$

Thus, we'll search for p value 0.8 and see what value of z is obtained.

The z value for p value 0.8 is z = 0.85

Thus,

$z = 0.85 = \dfrac{x - 420}{24}\\\\x = 0.85 \times 24 + 420 = 440.4$

• b)  Middle 60%

Let there are two values x and y such that :

$(x < X < y) = 60\% \: of \: values\\\\(X < y) + 1 - (X < x) = 60\% \: of \: values$

Since normal distribution is bell shaped, thus, for middle 60%, the rest 40% will lie 20% on left and 20% on right.

Since, in right, the limit starts from x= 440.4

thus, in left, the limit will start (to the left) from $420 - (440.4 - 420) = 399.6$

Thus,

For x = 399.4, and y = 440.4, the values in between are 60% of total values.

• c) Highest 80%

Since x = 399.4 to left is 20% of the data, thus, on its right lies 80% of the highest data, thus,

For $x > 399.4$ , we have 80% of data there.

• d) Lowest 15%

Let we have x such that for X < x, there lies 15% of data,

then using z score, we get:

$P(Z < \dfrac{x - 420}{24}) = 0.15$

From z tables, we get p value as z = -1.035

Thus,

$z = -1.035 = \dfrac{x - 420}{24}\\\\x = 420 - 1.305 \times 24 = 395.16$

The weight corresponding to each event is given as:

• a) Highest 20% : X > 440.4 grams,
• b Middle 60% : 399.4 < X < 440.4 grams
• c) Highest 80% :  X > 399.4 grams
• d) Lowest 15% :  X < 395.16 grams

Learn more about standard normal distribution here:

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